Problem: Let $f$ be a transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} -1 & 0 & -1 \\ \\ 3 & 0 & 0 \\ \\ 0 & \cos(y) & 1 \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(0, \pi, 0)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely
Solution: The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} -1 & 0 & -1 \\ \\ 3 & 0 & 0 \\ \\ 0 & \cos(y) & 1 \end{bmatrix} \right) \\ \\ &= -1 (0 \cdot 1 - 0 \cdot \cos(y)) \\ \\ &- 0 (3 \cdot 1 - 0 \cdot 0) \\ \\ &+ (-1)(3 \cdot \cos(y) - 0 \cdot 0) \\ \\ &= -3\cos(y) \end{aligned}$ If we evaluate $|J(f)|$ at $(0, \pi, 0)$, we get $3$. Because the Jacobian determinant here has an absolute value greater than $1$, we can conclude that $f$ will finitely expand the space around $(0, \pi, 0)$. To recap, the Jacobian determinant of $f$ is $-3 \cos(y)$, and $f$ will finitely expand the space around the point $(0, \pi, 0)$.